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Is there a close form for the following nested radical?

$$\sqrt{1+\sqrt[2!]{2^2+\sqrt[3!]{3^3+...}}}$$

It converges and
$$\quad \quad \lim_{n \to\infty} \sqrt{1+\sqrt[2!]{2^2+\sqrt[3!]{3^3+...+\sqrt[n!]{n^n}}}}=1.8430759846682...$$

Is this number algebraic or transcendental?

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    $\begingroup$ If it has a closed form, then $e$ is likely involved. $\endgroup$ Jul 2, 2015 at 21:21
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    $\begingroup$ I honestly doubt that a closed form will be found, but of course you never know. $\endgroup$
    – Wojowu
    Jul 2, 2015 at 21:30
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    $\begingroup$ Is there any context from which the above constant arises? If this is just a radical made on a spot, I see no reason for it to have a closed form. $\endgroup$
    – Wojowu
    Jul 2, 2015 at 21:39
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    $\begingroup$ I conjecture that almost surely 1) it does not have a closed form 2) it is trancendental 3) nobody will be able to provide an answer (proof) in one way or another to either of the two questions. $\endgroup$
    – Winther
    Jul 2, 2015 at 21:49

1 Answer 1

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Given the fact that neither Somos' quadratic recurrence constant, nor the nested radical constant are known to possess a closed form, I find it highly doubtful that this one will fare any better... Same as to the nature of the number, given that the nature of the afore-mentioned two constants is also unknown.

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  • $\begingroup$ See mathematica.stackexchange.com/questions/87359/… for numerical value obtained using Mathematica $\endgroup$
    – QuantumDot
    Nov 27, 2015 at 10:16
  • $\begingroup$ @QuantumDot: Never mind, I forgot to extract the square root. My sequence started $\sqrt[1!]{1^1+\sqrt[2!]{2^2+\ldots}}$ $\endgroup$
    – Lucian
    Nov 27, 2015 at 10:44

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